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Section 5.5 Other Options for Finding Algebraic Antiderivatives

We have learned two antidifferentiation techniques: \(u\)-substitution and integration by parts. The former is used to reverse the chain rule, while the latter to reverse the product rule. But we have seen that each works only in specialized circumstances. For example, while \(\int xe^{x^2} \, dx\) may be evaluated by \(u\)-substitution and \(\int x e^x \, dx\) by integration by parts, neither method provides a route to evaluate \(\int e^{x^2} \, dx\text{,}\) and in fact an elementary algebraic antiderivative for \(e^{x^2}\) does not exist. No antidifferentiation method will provide us with a simple algebraic formula for a function \(F(x)\) that satisfies \(F'(x) = e^{x^2}\text{.}\)

In this section of the text, our main goals are to identify some classes of functions that can be antidifferentiated, and to learn some methods to do so. We should also recognize that there are many functions for which an algebraic formula for an antiderivative does not exist, and appreciate the role that computing technology can play in finding antiderivatives of other complicated functions.

Preview Activity 5.5.1.

For each of the indefinite integrals below, the main question is to decide whether the integral can be evaluated using \(u\)-substitution, integration by parts, a combination of the two, or neither. For integrals for which your answer is affirmative, state the substitution(s) you would use. It is not necessary to actually evaluate any of the integrals completely, unless the integral can be evaluated immediately using a familiar basic antiderivative.

  1. \(\int x^2 \sin(x^3) \, dx\text{,}\)  \(\int x^2 \sin(x) \, dx\text{,}\)   \(\int \sin(x^3) \, dx\text{,}\)   \(\int x^5 \sin(x^3) \, dx\)

  2. \(\int \frac{1}{1+x^2} \, dx\text{,}\)   \(\int \frac{x}{1+x^2} \, dx\text{,}\)   \(\int \frac{2x+3}{1+x^2} \, dx\text{,}\)  \(\int \frac{e^x}{1+(e^x)^2} \, dx\text{,}\)

  3. \(\int x \ln(x) \, dx\text{,}\)   \(\int \frac{\ln(x)}{x} \, dx\text{,}\)   \(\int \ln(1+x^2) \, dx\text{,}\)  \(\int x\ln(1+x^2) \, dx\text{,}\)

  4. \(\int x \sqrt{1-x^2} \, dx\text{,}\)   \(\int \frac{1}{\sqrt{1-x^2}} \, dx\text{,}\)   \(\int \frac{x}{\sqrt{1-x^2}}\, dx\text{,}\)  \(\int \frac{1}{x\sqrt{1-x^2}} \, dx\text{,}\)

Subsection 5.5.1 The Method of Partial Fractions

The method of partial fractions is used to integrate rational functions. It involves reversing the process of finding a common denominator.

Evaluate

\begin{equation*} \int \frac{5x}{x^2-x-2} \, dx\text{.} \end{equation*}
Solution

If we factor the denominator, we can see how \(R\) might be the sum of two fractions of the form \(\frac{A}{x-2} + \frac{B}{x+1}\text{,}\) so we suppose that

\begin{equation*} \frac{5x}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1} \end{equation*}

and look for the constants \(A\) and \(B\text{.}\)

Multiplying both sides of this equation by \((x-2)(x+1)\text{,}\) we find that

\begin{equation*} 5x = A(x+1) + B(x-2)\text{.} \end{equation*}

Since we want this equation to hold for every value of \(x\text{,}\) we can use insightful choices of specific \(x\)-values to help us find \(A\) and \(B\text{.}\) Taking \(x = -1\text{,}\) we have

\begin{equation*} 5(-1) = A(0) + B(-3)\text{,} \end{equation*}

so \(B = \frac{5}{3}\text{.}\) Choosing \(x = 2\text{,}\) it follows

\begin{equation*} 5(2) = A(3) + B(0)\text{,} \end{equation*}

so \(A = \frac{10}{3}\text{.}\) Thus,

\begin{equation*} \int \frac{5x}{x^2-x-2} \, dx = \int \frac{10/3}{x-2} + \frac{5/3}{x+1} \, dx\text{.} \end{equation*}

This integral is straightforward to evaluate, and hence we find that

\begin{equation*} \int \frac{5x}{x^2-x-2} \, dx = \frac{10}{3} \ln|x-2| + \frac{5}{3}\ln|x+1| + C\text{.} \end{equation*}

It turns out that we can use the method of partial fractions to rewrite any ratinal function \(R(x) = \frac{P(x)}{Q(x)}\) where the degree of the polynomial \(P\) is less than 1  the degree of \(Q\) as a sum of simpler rational functions of one of the following forms:

\begin{equation*} \frac{A}{x-c},\ \frac{A}{(x-c)^n},\ \frac{Ax+B}{x^2 + k},\ \text{or }\frac{Ax+B}{\left(x^2 + k\right)^n} \end{equation*}

where \(A\text{,}\) \(B\text{,}\) and \(c\) are real numbers, and \(k\) is a positive real number. Because we can antidifferentiate each of these basic forms, partial fractions enables us to antidifferentiate any rational function.

If the degree of \(P\) is greater than or equal to the degree of \(Q\text{,}\) long division may be used to write \(R\) as the sum of a polynomial plus a rational function where the numerator's degree is less than the denominator's.

A computer algebra system such as Maple, Mathematica, or WolframAlpha can be used to find the partial fraction decomposition of any rational function. In WolframAlpha, entering

partial fraction 5x/(x^2-x-2)

results in the output

\begin{equation*} \frac{5x}{x^2-x-2} = \frac{10}{3(x-2)} + \frac{5}{3(x+1)}\text{.} \end{equation*}

We will use technology to generate partial fraction decompositions of rational functions, and then evaluate the integrals using established methods.

Activity 5.5.2.

For each of the following problems, evaluate the integral by using the partial fraction decomposition provided.

  1. \(\int \frac{1}{x^2 - 2x - 3} \, dx\text{,}\) given that \(\frac{1}{x^2 - 2x - 3} = \frac{1/4}{x-3} - \frac{1/4}{x+1}\)

  2. \(\int \frac{x^2+1}{x^3 - x^2} \, dx\text{,}\) given that \(\frac{x^2+1}{x^3 - x^2} = -\frac{1}{x} - \frac{1}{x^2} + \frac{2}{x-1}\)

  3. \(\int \frac{x-2}{x^4 + x^2}\, dx\text{,}\) given that \(\frac{x-2}{x^4 + x^2} = \frac{1}{x} - \frac{2}{x^2} + \frac{-x+2}{1+x^2}\)

Hint

  1. \(\int \ln(u) \, du = \frac{1}{u} + C\text{.}\)

  2. \(\frac{1}{x^2} = x^{-2}\text{.}\)

  3. \(\frac{-x+2}{1+x^2} = \frac{-x}{1+x^2}+\frac{2}{1+x^2}\text{,}\)

Answer

  1. \(\int \frac{1}{x^2 - 2x - 3} \, dx = \frac{1}{4}\ln|x-3| - \frac{1}{4}\ln|x+1| + C\text{.}\)

  2. \(\int \frac{x^2+1}{x^3 - x^2} \, dx = -\ln(x) + x^{-1} + 2\ln(x-1) + C\text{.}\)

  3. \(\int \frac{x-2}{x^4 + x^2}\, dx = \ln|x| + 2x^{-1} - \frac{1}{2} \ln(1+x^2) + 2\arctan(x) + C\text{.}\)

Solution

  1. \(\int \frac{1}{x^2 - 2x - 3} \, dx = \int \frac{1/4}{x-3} - \frac{1/4}{x+1} \, dx = \frac{1}{4}\ln|x-3| - \frac{1}{4}\ln|x+1| + C\text{.}\)

  2. \(\int \frac{x^2+1}{x^3 - x^2} \, dx = \int\left(-\frac{1}{x} - \frac{1}{x^2} + \frac{2}{x-1} \right) \, dx = -ln(x) + x^{-1} + 2\ln(x-1) + C\text{.}\)

  3. By the partial fractions decomposition, \(\int \frac{x-2}{x^4 + x^2}\, dx = \int \frac{1}{x} - \frac{2}{x^2} + \frac{-x+2}{1+x^2} \, dx \text{.}\) Observing that \(\frac{-x+2}{1+x^2} = \frac{-x}{1+x^2}+\frac{2}{1+x^2}\text{,}\) it now follows that

    \begin{equation*} \int \frac{x-2}{x^4 + x^2}\, dx = \int \frac{1}{x} - \frac{2}{x^2} - \frac{x}{1+x^2} + \frac{2}{1+x^2} \, dx\text{.} \end{equation*}

    Noting that \(\int \frac{x}{1+x^2} \, dx\) can be evaluated by the \(u\)-substitution \(u=1+x^2\text{,}\) we see that \(\int \frac{x}{1+x^2} \, dx = \frac{1}{2} \ln(1+x^2) + C\text{.}\) Thus

    \begin{equation*} \int \frac{x-2}{x^4 + x^2}\, dx = \int \frac{1}{x} - \frac{2}{x^2} - \frac{x}{1+x^2} + \frac{2}{1+x^2} \, dx = \ln|x| + 2x^{-1} - \frac{1}{2} \ln(1+x^2) + 2\arctan(x) + C\text{.} \end{equation*}

Subsection 5.5.2 Using an Integral Table

Calculus has a long history, going back to Greek mathematicians in 400-300 BC. Its main foundations were first investigated and understood independently by Isaac Newton and Gottfried Wilhelm Leibniz in the late 1600s, making the modern ideas of calculus well over 300 years old. It is instructive to realize that until the late 1980s, the personal computer did not exist, so calculus (and other mathematics) had to be done by hand for roughly 300 years. In the 21st century, however, computers have revolutionized many aspects of the world we live in, including mathematics. In this section we take a short historical tour to precede discussing the role computer algebra systems can play in evaluating indefinite integrals. In particular, we consider a class of integrals involving certain radical expressions.

As seen in the short table of integrals found in Appendix A, there are many forms of integrals that involve \(\sqrt{a^2 \pm w^2}\) and \(\sqrt{w^2 - a^2}\text{.}\) These integral rules can be developed using a technique known as trigonometric substitution that we choose to omit; instead, we will simply accept the results presented in the table. To see how these rules are used, consider the differences among

\begin{equation*} \int \frac{1}{\sqrt{1-x^2}} \,dx, \ \ \ \int \frac{x}{\sqrt{1-x^2}} \,dx, \ \ \ \text{and} \ \ \ \int \sqrt{1-x^2} \,dx\text{.} \end{equation*}

The first integral is a familiar basic one, and results in \(\arcsin(x) + C\text{.}\) The second integral can be evaluated using a standard \(u\)-substitution with \(u = 1-x^2\text{.}\) The third, however, is not familiar and does not lend itself to \(u\)-substitution.

In Appendix A, we find the rule

\begin{equation*} (h) ~ \int \sqrt{a^2 - u^2} \, du = \frac{u}{2}\sqrt{a^2 - u^2} + \frac{a^2}{2} \arcsin \frac{u}{a} + C\text{.} \end{equation*}

Using the substitutions \(a = 1\) and \(u = x\) (so that \(du = dx\)), it follows that

\begin{equation*} \int \sqrt{1-x^2} \, dx = \frac{x}{2} \sqrt{1-x^2} - \frac{1}{2} \arcsin x + C\text{.} \end{equation*}

Whenever we are applying a rule in the table, we are doing a \(u\)-substitution, especially when the substitution is more complicated than setting \(u = x\) as in the last example.

Evaluate the integral

\begin{equation*} \int \sqrt{9 + 64x^2} \, dx\text{.} \end{equation*}
Solution

Here, we want to use Rule (c) from the table, but we now set \(a = 3\) and \(u = 8x\text{.}\) We also choose the “\(+\)” option in the rule. With this substitution, it follows that \(du = 8dx\text{,}\) so \(dx = \frac{1}{8} du\text{.}\) Applying the substitution,

\begin{equation*} \int \sqrt{9 + 64x^2} \, dx = \int \sqrt{9 + u^2} \cdot \frac{1}{8} \, du = \frac{1}{8} \int \sqrt{9+u^2} \, du\text{.} \end{equation*}

By Rule (c), we now find that

\begin{align*} \int \sqrt{9 + 64x^2} \, dx =\mathstrut \amp \frac{1}{8} \left( \frac{u}{2}\sqrt{u^2 + 9} + \frac{9}{2}\ln|u + \sqrt{u^2 + 9}| + C \right)\\ =\mathstrut \amp \frac{1}{8} \left( \frac{8x}{2}\sqrt{64x^2 + 9} + \frac{9}{2}\ln|8x + \sqrt{64x^2 + 9}| + C \right)\text{.} \end{align*}

Whenever we use a \(u\)-subsitution in conjunction with Appendix A, it's important that we not forget to address any constants that arise and include them in our computations, such as the \(\frac{1}{8}\) that appeared in Example 5.5.2.

Activity 5.5.3.

For each of the following integrals, evaluate the integral using \(u\)-substitution and/or an entry from the table found in Appendix A.

  1. \(\int \sqrt{x^2 + 4} \, dx\)

  2. \(\int \frac{x}{\sqrt{x^2 +4}} \, dx\)

  3. \(\int \frac{2}{\sqrt{16+25x^2}}\, dx\)

  4. \(\int \frac{1}{x^2 \sqrt{49-36x^2}} \, dx\)

Hint

  1. Compare \(\int \sqrt{u^2 + a^2} \, du\text{.}\)

  2. Try a straightforward \(u\)-substitution; the table is unneeded.

  3. Let \(a = 4\) and \(u = 5x\) and look for a similar integral in the table.

  4. Let \(a = 7\) and \(u = 6x\text{;}\) find a related integral in the table.

Answer

  1. \(\int \sqrt{x^2 + 4} \, dx = \frac{1}{2} \arctan(\frac{x}{2})+C\text{.}\)

  2. \(\int \frac{x}{\sqrt{x^2 +4}} \, dx = \sqrt{x^2 + 4} + C\text{.}\)

  3. \(\int \frac{2}{\sqrt{16+25x^2}}\, dx = \frac{2}{5} \ln| 5x + \sqrt{16+25x^2} | + C\text{.}\)

  4. \(\int \frac{1}{x^2 \sqrt{49-36x^2}} \, dx = - \frac{\sqrt{49-36x^2}}{49x} + C\text{.}\)

Solution

  1. By (1) in Appendix A with \(a=2\) and \(u = x\text{,}\)

    \begin{equation*} \int \sqrt{x^2 + 4} \, dx = \frac{1}{2} \arctan(\frac{x}{2})+C\text{.} \end{equation*}

  2. Let \(u=x^2 + 4\text{,}\) so \(du = 2x dx\text{.}\) Thus

    \begin{equation*} \int \frac{x}{\sqrt{x^2 +4}} \, dx = \frac{1}{2} \int \frac{du}{\sqrt{u}} = \frac{1}{2} \cdot 2u^{1/2} + C = \sqrt{x^2 + 4} + C\text{.} \end{equation*}

  3. Letting \(a = 4\) and \(u = 5x\text{,}\) we see \(du=5dx\text{.}\) Thus,

    \begin{equation*} \int \frac{2}{\sqrt{16+25x^2}}\, dx = \frac{2}{5} \int \frac{du}{\sqrt{a^2 + u^2}}\text{.} \end{equation*}

    Now by (2) in Appendix A, it follows

    \begin{equation*} \int \frac{2}{\sqrt{16+25x^2}}\, dx = \frac{2}{5} \ln| 5x + \sqrt{16+25x^2} | + C\text{.} \end{equation*}

  4. Letting \(a = 7\) and \(u = 6x\text{,}\) it follows \(x=\frac{1}{6}u\) and \(du=6dx\text{,}\) and therefore

    \begin{equation*} \int \frac{1}{x^2 \sqrt{49-36x^2}} \, dx = \frac{1}{6} \int \frac{1}{\frac{1}{36}u^2 \sqrt{a^2-u^2}} \, du\text{.} \end{equation*}

    Using (11) in Appendix A and the fact that \(\frac{1}{6} \cdot 36 = 6\text{,}\) we see

    \begin{equation*} \int \frac{1}{x^2 \sqrt{49-36x^2}} \, dx = -6\cdot \frac{\sqrt{49-36x^2}}{49 \cdot 6x} + C\text{.} \end{equation*}

Subsection 5.5.3 Using Computer Algebra Systems

A computer algebra system (CAS) is a computer program that is capable of executing symbolic mathematics. For example, if we ask a CAS to solve the equation \(ax^2 + bx + c = 0\) for the variable \(x\text{,}\) where \(a\text{,}\) \(b\text{,}\) and \(c\) are arbitrary constants, the program will return \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\text{.}\) Research to develop the first CAS dates to the 1960s, and these programs became publicly available in the early 1990s. Two prominent examples are the programs Maple and Mathematica, which were among the first computer algebra systems to offer a graphical user interface. Today, Maple and Mathematica are exceptionally powerful professional software packages that can execute an amazing array of sophisticated mathematical computations. They are also very expensive, as each is a proprietary program. The CAS SAGE is an open-source, free alternative to Maple and Mathematica.

For the purposes of this text, when we need to use a CAS, we are going to turn instead to a similar, but somewhat different computational tool, the web-based “computational knowledge engine” called WolframAlpha. There are two features of WolframAlpha that make it stand out from the CAS options mentioned above: (1) unlike Maple and Mathematica, WolframAlpha is free (provided we are willing to navigate some pop-up advertising); and (2) unlike any of the three, the syntax in WolframAlpha is flexible. Think of WolframAlpha as being a little bit like doing a Google search: the program will interpret what is input, and then provide a summary of options.

If we want to have WolframAlpha evaluate an integral for us, we can provide it syntax such as

integrate x^2 dx

to which the program responds with

\begin{equation*} \int x^2 \, dx = \frac{x^3}{3} + \text{constant}\text{.} \end{equation*}

While there is much to be enthusiastic about regarding CAS programs such as WolframAlpha, there are several things we should be cautious about: (1) a CAS only responds to exactly what is input; (2) a CAS can answer using powerful functions from very advanced mathematics; and (3) there are problems that even a CAS cannot do without additional human insight.

Although (1) likely goes without saying, we have to be careful with our input: if we enter syntax that defines the wrong function, the CAS will work with precisely the function we define. For example, if we are interested in evaluating the integral

\begin{equation*} \int \frac{1}{16-5x^2} \, dx\text{,} \end{equation*}

and we mistakenly enter

integrate 1/16 - 5x^2 dx

a CAS will (correctly) reply with

\begin{equation*} \frac{1}{16}x - \frac{5}{3} x^3\text{.} \end{equation*}

But if we are sufficiently well-versed in antidifferentiation, we will recognize that this function cannot be the one that we seek: integrating a rational function such as \(\frac{1}{16-5x^2}\text{,}\) we expect the logarithm function to be present in the result.

Regarding (2), even for a relatively simple integral such as \(\int \frac{1}{16-5x^2} \, dx\text{,}\) some CASs will invoke advanced functions rather than simple ones. For instance, if we use Maple to execute the command

int(1/(16-5*x^2), x);

the program responds with

\begin{equation*} \int \frac{1}{16-5x^2} \, dx = \frac{\sqrt{5}}{20} \arctanh (\frac{\sqrt{5}}{4}x)\text{.} \end{equation*}

While this is correct (save for the missing arbitrary constant, which Maple never reports), the inverse hyperbolic tangent function is not a common nor familiar one; a simpler way to express this function can be found by using the partial fractions method, and happens to be the result reported by WolframAlpha:

\begin{equation*} \int \frac{1}{16-5x^2} \, dx = \frac{1}{8\sqrt{5}} \left(\log(4\sqrt{5}+5x) - \log(4\sqrt{5}-5x)\right) + \text{constant}\text{.} \end{equation*}

Using sophisticated functions from more advanced mathematics is sometimes the way a CAS says to the user “I don't know how to do this problem.” For example, if we want to evaluate

\begin{equation*} \int e^{-x^2} \, dx\text{,} \end{equation*}

and we ask WolframAlpha to do so, the input

integrate exp(-x^2) dx

results in the output

\begin{equation*} \int e^{-x^2} \, dx = \frac{\sqrt{\pi}}{2}\erf (x) + \text{constant}\text{.} \end{equation*}

The function “erf\((x)\)” is the error function, which is actually defined by an integral:

\begin{equation*} \erf (x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} \, dt\text{.} \end{equation*}

So, in producing output involving an integral, the CAS has basically reported back to us the very question we asked.

Finally, as remarked at (3) above, there are times that a CAS will actually fail without some additional human insight. If we consider the integral

\begin{equation*} \int (1+x)e^x \sqrt{1+x^2e^{2x}} \, dx \end{equation*}

and ask WolframAlpha to evaluate

int (1+x) * exp(x) * sqrt(1+x^2 * exp(2x)) dx,

the program thinks for a moment and then reports

(no result found in terms of standard mathematical functions)

But in fact this integral is not that difficult to evaluate. If we let \(u = xe^{x}\text{,}\) then \(du = (1+x)e^x \, dx\text{,}\) which means that the preceding integral has form

\begin{equation*} \int (1+x)e^x \sqrt{1+x^2e^{2x}} \, dx = \int \sqrt{1+u^2} \, du\text{,} \end{equation*}

which is a straightforward one for any CAS to evaluate.

So, we should proceed with some caution: while any CAS is capable of evaluating a wide range of integrals (both definite and indefinite), there are times when the result can mislead us. We must think carefully about the meaning of the output, whether it is consistent with what we expect, and whether or not it makes sense to proceed.

Subsection 5.5.4 Summary

  • We can antidifferentiate any rational function with the method of partial fractions. Any polynomial function can be factored into a product of linear and irreducible quadratic terms, so any rational function may be written as the sum of a polynomial plus rational terms of the form \(\frac{A}{(x-c)^n}\) (where \(n\) is a natural number) and \(\frac{Bx+C}{x^2 + k}\) (where \(k\) is a positive real number).

  • Until the development of compute algebra systems, integral tables enabled students of calculus to evaluate integrals such as \(\int \sqrt{a^2 + u^2} \, du\text{,}\) where \(a\) is a positive real number. A short table of integrals may be found in Appendix A.

  • Computer algebra systems can play an important role in finding antiderivatives, though we must be cautious to use correct input, to watch for unusual or unfamiliar advanced functions that the CAS may cite in its result, and to consider the possibility that a CAS may need further assistance or insight from us in order to answer a particular question.

Exercises 5.5.5 Exercises

1. Partial fractions: linear over difference of squares.
2. Partial fractions: constant over product.
3. Partial fractions: linear over quadratic.
4. Partial fractions: cubic over 4th degree.
5. Partial fractions: quadratic over factored cubic.
6.

For each of the following integrals involving rational functions, (1) use a CAS to find the partial fraction decomposition of the integrand; (2) evaluate the integral of the resulting function without the assistance of technology; (3) use a CAS to evaluate the original integral to test and compare your result in (2).

  1. \(\int \frac{x^3 + x + 1}{x^4 - 1} \, dx\)

  2. \(\int \frac{x^5 + x^2 + 3}{x^3 - 6x^2 + 11x - 6} \, dx\)

  3. \(\int \frac{x^2 - x - 1}{(x-3)^3} \, dx\)

Answer

Exercise Answer

Solution

  1. Wolfram|Alpha shows that

    \begin{equation*} \frac{x^3 + x + 1}{x^4 - 1} = -\frac{1}{2(x^2+1)} + \frac{1}{4(x+1)} + \frac{3}{4(x-1)}\text{.} \end{equation*}

    Then, using standard antiderivatives,

    \begin{align*} \int \frac{x^3 + x + 1}{x^4 - 1} \, dx &= \int -\frac{1}{2(x^2+1)} + \frac{1}{4(x+1)} + \frac{3}{4(x-1)} \ dx\\ &= \int -\frac{1}{2(x^2+1)} \ dx + \int \frac{1}{4(x+1)} \ dx + int \frac{3}{4(x-1)} \ dx\\ &= -\frac{1}{2} \arctan(x) + \frac{1}{4} \ln|x+1| + \frac{3}{4} \ln|x-1| + C \end{align*}

    Directly asking Wolfram|Alpha to integrate \(\frac{x^3 + x + 1}{x^4 - 1}\) yields

    \begin{equation*} \int \frac{x^3 + x + 1}{x^4 - 1} \, dx = \frac{1}{4} \left( 3\log(1-x) + \log(x+1) - 2\arctan{x} \right) + C\text{.} \end{equation*}

    By rearranging, distributing the \(\frac{1}{4}\text{,}\) and noticing that Wolfram|Alpha does not use absolute value with integrals involving the logarithm (as well as that \(|x-1| = |1-x|\)), we see that these two results are the same.

  2. Wolfram|Alpha shows that

    \begin{equation*} \frac{x^5 + x^2 + 3}{x^3 - 6x^2 + 11x - 6} = x^2 + 6x + 25 + \frac{255}{2(x-3)} - \frac{39}{x-2} + \frac{5}{2(x-1)}\text{.} \end{equation*}

    Then, using standard antiderivatives,

    \begin{align*} \int \frac{x^5 + x^2 + 3}{x^3 - 6x^2 + 11x - 6} \, dx &= \int x^2 + 6x + 25 + \frac{255}{2(x-3)} - \frac{39}{x-2} + \frac{5}{2(x-1)} \ dx\\ &= \int x^2 \ dx + \int 6x \ dx + \int 25 \ dx + \int \frac{255}{2(x-3)} \ dx - \int \frac{39}{x-2} \ dx + \int \frac{5}{2(x-1)} \ dx\\ &= \frac{x^3}{3} + 3x^2 + 25x + \frac{255}{2} \ln|x-3| - 39 \ln|x-2| + \frac{5}{2} \ln|x-1| + C\text{.} \end{align*}

    Directly asking Wolfram|Alpha to integrate \(\frac{x^5 + x^2 + 3}{x^3 - 6x^2 + 11x - 6}\) yields the same result with the caveats mentioned in part (a).

  3. Wolfram|Alpha shows that

    \begin{equation*} \frac{x^2 - x - 1}{(x-3)^3} = \frac{1}{x-3} + \frac{5}{(x-3)^2} + \frac{5}{(x-3)^3}\text{.} \end{equation*}

    Then, using standard antiderivatives,

    \begin{align*} \int \frac{x^2 - x - 1}{(x-3)^3} \, dx &= \int \frac{1}{x-3} + \frac{5}{(x-3)^2} + \frac{5}{(x-3)^3} \ dx\\ &= \int \frac{1}{x-3} \ dx + \int \frac{5}{(x-3)^2} \ dx + \int \frac{5}{(x-3)^3} \ dx\\ &= \ln|x-3| + \int 5(x-3)^{-2} \ dx + \int 5(x-3)^{-3} \ dx\\ &= \ln|x-3| - \frac{5}{x-3} - \frac{5}{2(x-3)^2} + C\text{.} \end{align*}

    Directly asking Wolfram|Alpha to integrate \(\frac{x^2 - x - 1}{(x-3)^3}\) yields

    \begin{equation*} \int \frac{x^2 - x - 1}{(x-3)^3} \, dx = \log(x-3) - \frac{5(2x-5)}{2(x-3)^2} + C\text{.} \end{equation*}

    The logarithm part is the same, and since

    \begin{equation*} -\frac{5}{x-3} - \frac{5}{2(x-3)^2} = -\frac{5[2(x-3) + 1]}{2(x-3)^2} = -\frac{5(2x-5)}{2(x-3)^2}\text{,} \end{equation*}

    we see that the rest of the result matches as well.

7.

For each of the following integrals involving radical functions, (1) use an appropriate \(u\)-substitution along with Appendix A to evaluate the integral without the assistance of technology, and (2) use a CAS to evaluate the original integral to test and compare your result in (1).

  1. \(\int \frac{1}{x \sqrt{9x^2 + 25}} \, dx\)

  2. \(\int x \sqrt{1 + x^4} \, dx\)

  3. \(\int e^x \sqrt{4 + e^{2x}} \, dx\)

  4. \(\int \frac{\tan(x)}{\sqrt{9 - \cos^2(x)}} \, dx\)

Answer

  1. \(\int \frac{1}{x \sqrt{9x^2 + 25}} \, dx = -\frac{1}{5} \ln \left| \frac{5+\sqrt{9x^2 + 5^2}}{3x} \right| + C\text{.}\)

  2. \(\int x \sqrt{1 + x^4} \, dx = \frac{1}{2} \left( \frac{x^2}{2}\sqrt{x^4 + 1} + \frac{1}{2}\ln|x^2 + \sqrt{x^4 + 1}| \right) + C\)

  3. \(\int e^x \sqrt{4 + e^{2x}} \, dx = \frac{e^x}{2}\sqrt{e^{2x} + 4} + 2\ln|e^x + \sqrt{e^{2x} + 4}| + C\)

  4. \(\int \frac{\tan(x)}{\sqrt{9 - \cos^2(x)}} \, dx = \frac{1}{3} \ln \left| \frac{3 + \sqrt{9 - \cos^2(x)}}{\cos(x)} \right| + C\text{.}\)

Solution

  1. For \(\int \frac{1}{x \sqrt{9x^2 + 25}} \, dx\text{,}\) we let \(u = 3x\) so \(du = 3 \, dx\) and \(x = \frac{1}{3} u\text{.}\) Applying this substitution,

    \begin{equation*} \int \frac{1}{x \sqrt{9x^2 + 25}} \, dx = \int \frac{3}{u \sqrt{u^2 + 25}} \cdot \frac{1}{3} \, du = \int \frac{1}{u \sqrt{u^2 + 25}} \, du\text{.} \end{equation*}

    Now, using Rule (e) from the appendix with \(a = 5\text{,}\)

    \begin{equation*} \int \frac{1}{u \sqrt{u^2 + 25}} \, du = -\frac{1}{5} \ln \left| \frac{5+\sqrt{u^2 + 5^2}}{u} \right| + C \end{equation*}

    Substituting back to \(x\text{,}\) we've shown that

    \begin{equation*} \int \frac{1}{x \sqrt{9x^2 + 25}} \, dx = -\frac{1}{5} \ln \left| \frac{5+\sqrt{9x^2 + 5^2}}{3x} \right| + C\text{.} \end{equation*}

    Asking Wolfram|Alpha to evaluate the same original integral, the program reports that

    \begin{equation*} \int \frac{1}{x \sqrt{9x^2 + 25}} \, dx = \frac{1}{5} \left( \log(x) - \log\left( \sqrt{9x^2 + 25} + 5\right) \right) + C\text{.} \end{equation*}

    These two results are in fact the same (the program uses ``\(\log\)'' where we use ``\(\ln\)''), though it takes some work with logarithm properties to show this. Using the rule \(\ln(\frac{u}{v}) = \ln (u) - \ln (v)\) and \(\ln(3x) = \ln(3) + \ln(x)\) and including the value \(\ln(3)\) in the arbitrary constant of integration, it follows that the two results match.

  2. For \(\int x \sqrt{1 + x^4} \, dx\text{,}\) we first recognize that \(x^4 = (x^2)^2\text{,}\) and that with \(u = x^2\) we have a function-derivative pair present with \(du = 2x \, dx\text{.}\) Applying this substitution,

    \begin{equation*} \int x \sqrt{1 + x^4} \, dx = \int \frac{1}{2} \sqrt{1+u^2} \, du\text{.} \end{equation*}

    Next, we use Rule (c) with ``\(+\)'' and \(a = 1\) to find that

    \begin{equation*} \int \frac{1}{2} \sqrt{1+u^2} \, du = \frac{1}{2} \left( \frac{u}{2}\sqrt{u^2 + 1} + \frac{1}{2}\ln|u + \sqrt{u^2 + 1}| \right) + C \end{equation*}

    Returning to the original variable, \(x\text{,}\) we have

    \begin{equation*} \int x \sqrt{1 + x^4} \, dx = \frac{1}{2} \left( \frac{x^2}{2}\sqrt{x^4 + 1} + \frac{1}{2}\ln|x^2 + \sqrt{x^4 + 1}| \right) + C \end{equation*}

    Wolfram|Alpha's first response when asked to evaluate this integral is to give a result that involves the hyperbolic sine function (\(\sinh^{-1}(x^2)\)); if we scroll down to ``alternate results'', we find one that matches what we've found above.

  3. For \(\int e^x \sqrt{4 + e^{2x}} \, dx\text{,}\) we first observe that \(e^{2x} = (e^x)^2\text{.}\) If we use the substitution \(u = e^x\text{,}\) the resulting integral is

    \begin{equation*} \int e^x \sqrt{4 + e^{2x}} \, dx = \int \sqrt{4 + u^2} \, du\text{,} \end{equation*}

    which can be evaluated by the same approach as in part (b) with Rule (c) from the Appendix, this time with \(a = 2\text{.}\) Thus,

    \begin{equation*} \int \sqrt{4 + u^2} \, du = \frac{u}{2}\sqrt{u^2 + 4} + \frac{2^2}{2}\ln|u + \sqrt{u^2 + 4}| \end{equation*}

    Substituting back,

    \begin{equation*} \int e^x \sqrt{4 + e^{2x}} \, dx = \frac{e^x}{2}\sqrt{e^{2x} + 4} + 2\ln|e^x + \sqrt{e^{2x} + 4}| + C \end{equation*}

    As in part (b), we need to scroll down in Wolfram|Alpha's results to find one close to what we have here. The alternate result almost matches, but the program writes

    \begin{equation*} \log(\left( \sqrt{\frac{e^{2x}}{4} + 1 } + \frac{e^{x}}{2} \right) \end{equation*}

    where we have

    \begin{equation*} \ln|e^x + \sqrt{e^{2x} + 4}.| \end{equation*}

    This discrepancy can be explained using properties of the logarithm function and the presence of the arbitrary constant. If we write

    \begin{equation*} \log(\left( \sqrt{\frac{e^{2x}}{4} + 1 } + \frac{e^{x}}{2} \right) = \log(\left( \sqrt{\frac{e^{2x} + 4}{4}} + \frac{e^{x}}{2} \right) = \log(\left( \frac{ \sqrt{e^{2x} + 4} + e^{x}}{2} \right) \end{equation*}

    and then recall that \(\log \left( \frac{z}{2} \right) = \log(z) - \log(2)\text{,}\) we see that the \(\log(2)\) can be included in the arbitray constant, and the two different results actually match.

  4. To evaluate the integral \(\int \frac{\tan(x)}{\sqrt{9 - \cos^2(x)}} \, dx\text{,}\) we first write \(\tan(x) = \frac{\sin(x)}{\cos(x)}\) to have the integral in the form

    \begin{equation*} \int \frac{\tan(x)}{\sqrt{9 - \cos^2(x)}} \, dx = \int \frac{\sin(x)}{\cos(x) \sqrt{9 - \cos^2(x)}} \, dx\text{.} \end{equation*}

    Next, letting \(u = \cos(x)\text{,}\) it follows \(du = -\sin(x) \, dx\text{,}\) and therefore

    \begin{equation*} \int \frac{\tan(x)}{\sqrt{9 - \cos^2(x)}} \, dx = \int \frac{-1}{u \sqrt{9 - u^2}} \, du\text{.} \end{equation*}

    By Rule (j) in the appendix, we know that

    \begin{equation*} \int \frac{1}{u \sqrt{9 - u^2}} \, du = -\frac{1}{3} \ln \left| \frac{3 + \sqrt{9 - u^2}}{u} \right| + C\text{.} \end{equation*}

    Connecting the above chain of equalities and substituting back to \(x\text{,}\) we have shown that

    \begin{equation*} \int \frac{\tan(x)}{\sqrt{9 - \cos^2(x)}} \, dx = \frac{1}{3} \ln \left| \frac{3 + \sqrt{9 - \cos^2(x)}}{\cos(x)} \right| + C\text{.} \end{equation*}

    When Wolfram|Alpha is asked to evaluate the original integral, its first result involves the inverse hyperbolic tangent function, and the alternate forms are equally complicated. It seems that the technology is unable to evaluate this integral in the most straightforward way.

8.

Consider the indefinite integral given by

\begin{equation*} \int \frac{\sqrt{x+\sqrt{1+x^2}}}{x} \, dx\text{.} \end{equation*}

  1. Explain why \(u\)-substitution does not offer a way to simplify this integral by discussing at least two different options you might try for \(u\text{.}\)

  2. Explain why integration by parts does not seem to be a reasonable way to proceed, either, by considering one option for \(u\) and \(dv\text{.}\)

  3. Is there any line in the integral table in Appendix A that is helpful for this integral?

  4. Evaluate the given integral using WolframAlpha. What do you observe?

Answer

  1. Try \(u = 1+x^2\) or \(u = x + \sqrt{1+x^2}\text{.}\)

  2. Try \(u = \sqrt{x+\sqrt{1+x^2}}\) and \(dv = \frac{1}{x} \, dx\text{.}\)

  3. No.

  4. It appears that the function \(\frac{\sqrt{x+\sqrt{1+x^2}}}{x}\) does not have an elementary antiderivative.

Solution

  1. Looking at the composite functions that are present, there are two possible \(u\)-substitutions to try. One is \(u = 1+x^2\text{,}\) for which \(du = 2x \, dx\) and \(x = \sqrt{u-1}\text{.}\) Because there is no \(x \, dx\) term present in the integral, this substitution will not work.

    A second subsitution to attempt is letting \(u = x + \sqrt{1+x^2}\text{,}\) for which \(du = \left(1 + x(1+x^2)^{-1/2} \right) \, dx\text{.}\) Here, too, the needed product is not present in the integral in order for the subsitution to work.

  2. For integration by parts to work, we normally need a choice of \(u\) that is straightforward to differentiate, and \(dv\) that is straightforward to antidifferentiate. Noting that the integrand can really only be written as the product of \(\frac{1}{x}\) and \(\sqrt{x+\sqrt{1+x^2}}\text{,}\) it appears that the most natural possible option is to let \(u = \sqrt{x+\sqrt{1+x^2}}\) and \(dv = \frac{1}{x} \, dx\text{,}\) since we can antidifferentiate \(\frac{1}{x}\text{.}\) Here, however, the approach stalls, because integration by parts requires us to evaluate \(\int v \, du\text{,}\) and that integral is

    \begin{equation*} \int \ln|x| \cdot \frac{1}{2} \left( x + \sqrt{1+x^2} \right)^{-1/2} \cdot \left( 1 + x(1+x^2)^{-1/2} \right) \, dx \end{equation*}

    which is far more complicated than the already-complicated integral we started with. From this it appears that integration by parts will not be productive.

  3. Since we can't seem to find a \(u\)-substitution that simplifies the given integral, and there are no formulas in the Appendix that use nested square roots, it appears that the table doesn't offer any help in evaluating this particular integral.

  4. Wolfram|Alpha returns results that either involve inverse hyperbolic trigonometric functions or complex numbers. As such, it appears that the function \(\frac{\sqrt{x+\sqrt{1+x^2}}}{x}\) does not have an elementary antiderivative that we can find.

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